作者b94501010 (海鲜不新鲜)
看板trans_math
标题Re: [积分] 关於弧长的问题??
时间Sat Aug 25 16:57:05 2007
※ 引述《gn00067504 (ㄚ良)》之铭言：
: Find the arc length of the curve r=1-cos西塔 0<西塔<180度
: 拜托大家帮个忙了谢谢....
π dr
L = ∫ √[r^2 + (----)^2] dθ
0 dθ
π
= ∫ √[(1-cosθ)^2 + (sinθ)^2] dθ
0
π
= ∫ √(2-2cosθ) dθ
0
π
= √2 ∫ √(1-cosθ) dθ
0
Let u = √(1-cosθ) => u^2 = 1-cosθ
=> cosθ = 1-u^2
=> sinθ = √1-(1-u^2)^2 since 0 <θ< π
= u √(2-u^2)
2u 2
Then 2udu = sinθdθ => dθ = ------------- du = ----------- du
u √(2-u^2) √(2-u^2)
When θ= 0, u = 0 ; when θ= π, u = √2
π
Thus, L = √2 ∫ √(1-cosθ) dθ
0
√2 2
= √2 ∫ u ----------- du
0 √(2-u^2)
√2
= 2 √2 [- √(2-u^2) ]
0
= 4
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